Duration 25 min 1(a) The following data refer to the ions of the element vanadium in acidic solution. aqueous ions Yellow Blue Green Violet

Transcription

1 Name/CG: JC2 H2 Chemistry Electrochemistry Revision Duration 25 min 1(a) The following data refer to the ions of the element vanadium in acidic solution. Ions + VO 2 VO 2+ V 3+ V 2+ Colour of aqueous ions Yellow Blue Green Violet For each of the following situations, use E o cell data to predict what might happen when the two reagents are mixed together. Write balanced equations for any reactions that occur. excess Zn(s) and VO 2+ (aq) Starting reagent is VO 2+ and Zn To check if Zn can reduce VO 2+. VO H + + e - V 3+ + H 2 O E θ = +0.34V Zn e -. Zn E θ = -0.76V E θ cell = E θ red - E θ oxid = (-0.76) = +1.10V [1/2] Since E θ cell > 0, reaction is feasible. Zn(s) + 2VO 2+ (aq) + 4H + (aq) Zn 2+ (aq) + 2V 3+ (aq) + 2H 2 O (l) [1/2] Colour change of solution from blue to green. [1/2] To check if Zn can reduce V 3+. V 3+ + e -. V 2+ E = -0.26V θ Zn e -. Zn E = θ -0.76V E θ cell = E θ red - E θ oxid = (-0.76) = +0.50V [1/2] Since E θ cell > 0, reaction is feasible. Zn(s) + 2V 3+ (aq) Zn 2+ (aq) + 2V 2+ (aq) [1/2] 1

2 Colour change of solution from green to violet. [1/2] In conclusion, Zn(s) + VO 2+ (aq) + 2H + (aq) Zn 2+ (aq) + V 2+ (aq) + H 2 O (l) Zn can reduce VO 2+ to V 2+. Solution changes from blue to green and finally violet in colour. (ii) S 4 O 6 2- (aq) and V 2+ (aq) [4] S 4 O e - 2-2S 2 O (Reduction) V 3+ + e -. V (Oxidation) E θ cell = E θ red - E θ oxid = 0.09 (-0.26) = +0.35V [1/2] Since E θ cell > 0, reaction is feasible. Solution changes from violet to green. [1/2] S 4 O 6 2- (aq) + 2V 2+ (aq) 2S 2 O 3 2- (aq) + 2V 3+ (aq) [1/2] 2

3 (b) The magnitudes (but not the sign) of the standard electrode potentials of two metals B and C are given below: Equation Magnitude of E o B e B 0.30 C e C 0.36 When B 2+ /B is connected to the standard hydrogen electrode, electrons flow from the B electrode to the platinum electrode of the standard hydrogen electrode. When the half-cells of B 2+ /B and C 2+ /C are connected, electrons flow from the B electrode to the C electrode. Deduce the signs of the E o values of the B 2+ /B and C 2+ /C half-cells. Explain your answer. Since electrons flow from the B electrode (i.e anode/oxidation) [1/2] to the platinum electrode of SHE and E θ cell = E θ red - E θ oxid> 0 OR 0.00 ( 0.30) > 0 [1/2] sign of the E θ value of the B 2+ /B is negative [1/2] Since electrons flow from the B electrode (i.e anode/oxidation) to the C electrode and E θ cell = E θ red - E θ oxid> ( 0.30) > 0 [1/2] sign of the E θ value of the C 2+ /C is positive [1/2] or other logical reasoning / working (ii) Calculate the e.m.f. of the cell made up of B 2+ /B and C 2+ /C half-cells. emf of cell = ( 0.30) = V [1] no units / + sign minus ½ (iii) What would be the effect on the e.m.f. of the cell in (b)(ii) when the concentration of C 2+ is decreased. C e C When [C 2+ ] decreased, the equilibrium position shifts to the left [1/2] to increase [C 2+ ] by Le Chatelier s Principle. no mark if eqm is not written Oxidation is more likely to occur. Electrode potential E decreases [1/2] θ Hence E θ cell = E θ - E θ decreases [1]. 3

4 (iv) Suggest a replacement half-cell for B 2+ /B half-cell which would reverse the direction of electron flow in the C 2+ /C half-cell. [8] E θ cell = E θ red > 0 Choose half-cell with E θ > 0.36V Ag + / Ag [1] or other correct half-cell (c) Explain the following observations as fully as you can. The electrolysis of dilute potassium bromide chloride produces oxygen gas at the anode, whereas the electrolysis of concentrated potassium bromide chloride results in the formation of a orange-red a yellowish-green colouration at the anode region. [4] Cl 2 + 2e 2Cl - E = V 4H + (l) + O 2 (g) + 4e 2 H 2 O (l) E = V For dilute potassium bromide: Both Cl - and H 2 O migrate to anode H 2 O is preferentially discharged as it has a less positive E θ [1/2] value compared to Cl - Oxidation: 2 H 2 O (l) 4H + (l) + O 2 (g) + 4e [1/2] Hence effervescence of O 2 gas is observed For concentrated potassium bromide: Both Cl - and H 2 O migrate to anode Cl - is preferentially discharged even though it is has a more positive E θ value compared to H 2 O because the concentration of Cl - is very high. [1/2] Oxidation: 2Cl - (aq) Cl 2 (aq) + 2e [1/2] Hence effervescence of pale yellow due to Cl 2 is observed. 4

5 RJC Prelim 05/P2/Q2 (18 min) 2 Aluminium/air batteries are used as back-up power supplies in many telephone exchanges. The structure of one such battery is shown below: porous electrode Z terminal X aluminium alloy terminal Y electrolyte The electrolyte used is aqueous potassium hydroxide. In the reaction, oxygen in air is reduced while aluminium is oxidised as shown by the following equation: Al (s) + 4OH (aq) Al(OH) 4 (aq) + 3e (a) Is terminal Y negative or positive? Give a reason for your answer. [2] Terminal Y is negative. This is because electrons are produced by the oxidation of aluminium and leave the battery via terminal Y. (b) Write a balanced equation with state symbols for the reaction that takes place at the porous electrode Z. [1] O 2 (g) + 2H 2 O (l) + 4e 4OH (aq) (c) Why must electrode Z be porous? [1] Z must be porous so that oxygen may diffuse through the electrode and come into contact with the electrolyte where it is reduced. (d) Sometimes, electrode Z becomes clogged up so that it is no longer porous. This causes the aluminium/air battery to be potentially explosive. With the help of a suitable equation, explain why this is so. [2] Instead of oxygen being reduced, water may be reduced at electrode Z forming hydrogen gas: 2H 2 O (l) + 2e H 2 (g) + 2OH (aq) H 2 gas is produced in a sealed container, so potentially explosive. (e) The aluminium used is specially alloyed so that it is not easily oxidised by air. Explain why ordinary aluminium cannot be used. [1] 5

6 Ordinary aluminium is protected by a dense, impervious layer of aluminium oxide. This layer prevents aluminium from being oxidised so that it cannot act as an anode. (f) Besides aluminium, zinc may also be used. Give two advantages of using aluminium rather than zinc. [2] Zn e Zn E O = 0.76V Al e Al E O = 1.66V Aluminium has a more negative standard reduction potential/ bigger standard oxidation potential than zinc so that the operating voltage of aluminium/air battery is higher than zinc/ air battery. Aluminium produces more electron per mole of metal so that more electrons can be produced from aluminium/air battery than zinc/ air battery for the same mass of metal used. Aluminium is lighter than zinc (Accept other sensible answers, but not aluminium is cheaper than Zn, less corrosive than Zn) (g) When aqueous potassium hydroxide is replaced by aqueous sodium chloride, it is noticed that a white solid is formed around the aluminium alloy electrode. Name the white solid and explain why it is formed. [2] The white solid is aluminium hydroxide. There are insufficient hydroxide ions to dissolve the insoluble amphoteric aluminium hydroxide. Al(OH) 3 is formed due to the precipitation reaction between Al 3+ (produced by oxidation of Al) and OH (produced by reduction of oxygen.) (h) The aluminium/air battery with aqueous potassium hydroxide as an electrolyte was used as an electrical source to electroplate an article with copper. If the mass of aluminium electrode decreased by 5.4 g, calculate the mass of copper that had plated out of the solution. [2] Al 3e Cu 2+ (aq) + 2e Cu (s) Hence 2 Al 3 Cu No.of mol of Al used = 5.4/27 = 0.2 No of mol of Cu formed = 3/2 x 0.2 = Mass of Cu = x 63.5 g = 19.1 g 6

7 NJC Prelim 05/P2/Q2 (15 min) 3(a) Iron is the most widely used d-block metal in the construction industry, mainly because it is very abundant in the earth s crust. However, we are also familiar with the red-brown solid known as rust, which weakens the structural strength of iron. The control of corrosion of iron is hence of great economic importance. The scheme below outlines the stages in the rusting process. prolonged Stage I exposure to air Fe(s) Fe(OH) 2 (s) Fe 2 O 3.xH 2 O(s) moist air red-brown rust Why do you think that the structural strength of iron is weakened by the formation of rust? Formation of rust, an ionic compound, will weaken the structural strength of iron, as rust is brittle. (ii) Stage I of rusting is an electrochemical process. air C water A iron Taking A to be the anode, and C to be the cathode of the electrochemical cell reaction of rusting in the above diagram, 1 write ion-electron equations for the process at A : Fe (s) Fe 2+ (aq) + 2e C : O 2 + 2H 2 O + 4e 4OH 2 By means of an arrow, show the direction of electron flow in the above cell. 3 Show, by a cross (X) on the diagram, one spot where you would expect most rust to be accumulated. 7

8 (iii) Use information from your Data Booklet, estimate a value for the potential of the cell in (a)(ii), giving reasons. E cell o = 0.44 (- 0.40) = +0.84V The value of the potential of the cell in (a)(ii) will be less than +0.84V as the conditions of the cell is not of standard conditions. [6] (b) Unlike iron, aluminium has resistance to atmospheric corrosion because of the oxide layer on its surface. State one common use of aluminium which makes use of this property. Cans for soft drinks, car bodies. (ii) This oxide layer can be thickened by an electrolytic process. On the space below, draw a fully-labelled diagram to show how this process can take place. [4] Graphite Cathode Aluminium Anode dil H 2 SO 4 (aq) 8

9 Homework Duration: 9 min 1 Predict the products formed at the anode, and at the cathode, when the following liquids are electrolyzed using inert electrodes. (a) NaBr(l) For electrolysis: At anode(+): Br- is oxidized to Br 2. Hence product is Br 2. At cathode(-): Na + is reduced to Na. Hence product is Na. (b) NaBr(aq) Species present in dilute sodium chloride: Na +, Br -, H 2 O At the anode (+) Both H 2 O and Br - migrate to anode Preferential oxidation of Br - as E Ө of Br 2 / Br - (+1.07 V) is less positive than E Ө of O 2 / H 2 O (+1.23V) Hence product is Br 2 At the cathode (-) Both H 2 O and Na + migrate to cathode Preferential reduction of H 2 O as E Ө of H 2 O / H 2 (-0.83 V) is more positive than E Ө of Na + / Na (-2.71V) Hence product is H 2 (c) CuF 2 (aq) Species present in dilute sodium chloride: Cu 2+, F -, H 2 O At the anode (+) Both H 2 O and F - migrate to anode Preferential oxidation of H 2 O as E Ө of O 2 / H 2 O (+1.23V) is less positive than E Ө of F 2 / F -- (+2.87 V) Hence product is O 2 At the cathode (-) Both H 2 O and Cu 2+ migrate to cathode Preferential reduction of Cu 2+ as E Ө of Cu 2+ / Cu (+0.34 V) is more positive than E Ө of H 2 O / H 2 (-0.83V) Hence product is Cu [6] 9

10 Duration: 6 min 2(a) By means of a fully labelled diagram, describe how the standard electrode potential of Cr 3+ /Cr 2+ system can be measured by using standard hydrogen electrode. [4] Voltmeter Pt (s) Cr 2+ ions (1 mol dm -3 ) & Cr 3+ ions (1 mol dm -3 ) [diagram [2], anything missing or wrong -1/2 till zero] (b) Use the Data Booklet to predict the outcome of mixing acidified aqueous hydrogen peroxide with aqueous bromine, Br 2 is reduced to Br - H 2 O 2 is oxidised to O 2 E θ cell = = V [1/2] > 0 and reaction is feasible Br 2 (aq) + H 2 O 2 2Br - (aq) + O 2 (g) + 2H + (aq) [1] no state symbol minus ½ Reddish-brown aqueous bromine is decolourised [1/2] and effervescence of O 2 gas is seen [1/2]. (ii) aqueous iron(ii) sulfate. Include E o cell, balanced equation and observation in your answer. Fe 2+ is oxidised to Fe 3+ H 2 O 2 is reduced to H 2 O E θ cell = = V [1/2] > 0 and reaction is feasible 2Fe 2+ (aq) + H 2 O 2 + 2H + (aq) 2Fe 3+ (aq) + 2H 2 O(l) [1] no s.s. minus ½ Green Fe 2+ (aq) solution turns to yellow Fe 3+ (aq) [1/2]. [4] 10

11 Duration: 15 min 3(a) An electrochemical cell has an overall e.m.f. of V at 25 C. The overall cell reaction is shown below. The cell diagram is given as: 2CO (g) + O 2 (g) 2CO 2 (g) Ni (s) CO (g), CO 2 (g) CO 3 2- (aq) CO 3 2- (aq) O 2 (g), CO 2 (g) Ag (s) The reaction occurring at the silver electrode is: 2CO 2 (g) + O 2 (g) + 4e - 2CO 3 2- (aq) E θ = V Deduce the reaction occurring at the nickel electrode and calculate the standard electrode potential for the reaction involved. Anode: 2CO (g) + 2CO 3 2- (aq) 4CO 2 (g) + 4e - [1] E θ cell = E θ red E θ oxid = E θ oxid E θ oxid = = 0.64 V [1] (ii) Draw a labelled diagram to show the laboratory experimental set-up of the electrochemical cell under standard conditions. Indicate the direction of the electron flow in your sketch. Electron flow V CO 2 (g) and CO (g) at 298K and 1atm salt bridge CO 2 (g) and O 2 (g) at 298K and 1atm Ni electrode [CO 3 2- (aq)] = 1moldm -3 Ag electrode [CO 3 2- (aq)] = 1moldm -3 [3m, anything missing minus ½] 11

12 (iii) Suggest a replacement half cell, involving a gas, for CO 3 2- (aq) O 2 (g), CO 2 (g) Ag (s), which would reverse the direction of the electron flow in the CO 3 2- (aq) CO 2 (g), CO (g) Ni(s) half cell. Your answer needs to state both the electrode and the reagents of your new half-cell. [6] To reverse electron flow, the selected half cell should now undergo oxidation & the other half cell with the Ni electrode will undergo reduction (E θ red = 0.64 V) E θ cell = E θ red E θ oxid E θ cell = 0.64 E θ oxid > 0 E θ oxid < 0.64 V Pt (s) H 2 O (l) OH - (aq) H 2 (g) [2] Or Pt(s) [1] KOH(aq), H 2 (g) [1] (b) A modified version of the Leclanche cell is now widely used as a dry cell battery to power portable gadgets. Two possible half-cell reactions in the dry cell are: [Zn(NH 3 ) 4 ] 2+ (aq) + 2e - Zn (s) + 4NH 3 (aq) E θ = V NH 4 + (aq) + MnO 2 (s) + H 2 O (l) + e - Mn(OH) 3 (s) + NH 3 (aq) E θ = V Calculate the standard electrode potential, E θ of the dry cell. E θ cell = E θ red E θ oxid = 1.00 (-1.03) = V [1] 12

13 (ii) The actual e.m.f. of the cell is V. Give a reason as to why this value is different from the one calculated in (b). Due to non-standard conditions / non-standard concentrations. [1] (iii) Determine the amount of time in seconds required for a dry cell containing 6.50 g of MnO 2 to produce a constant current of A. [4] No. of mol of MnO 2 = 6.50 / 86.9 = mol [1] 1 mol of MnO 2 1 mol of e = t = s [1] 13